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MCAT Physical : Gas Laws

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Example Questions

Example Question #1 : Gas Laws

15L of a gas is held at constant pressure as the temperature increases from 300K to 350K. What is the new volume?

Possible Answers:

17.5L

12.5L

The volume does not change

10L

Correct answer:

17.5L

Explanation:

Charles's Law states that $\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}$ . To solve for the final volume, we simply plug in our given values to this equation.

$V_2=\frac{V_1T_2}{T_1}=\frac{(15L)(350K)}{(300K)}=17.5L$

This law applies only for isobaric (constant pressure) changes.

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Example Question #1 : Gas Laws

8 liters of an ideal gas is in an isolated container at 30 degrees Celsius. The container is heated at constant pressure until its volume is doubled. What is the new temperature of the gas?

Possible Answers:

150^oC

220^oC

333^oC

124^oC

60^oC

Correct answer:

333^oC

Explanation:

At constant pressure, $\frac{V_{1}}{T_{1}}=\frac{V_{2}}{T_{2}}$ , where the temperatures are measured in Kelvin (absolute temperature).

First, convert the given temperature (C) to Kelvin (K).

K = C + 273 = 30 + 273 = 303K

Plug the temperature and volumes into the above equation and solve for the final temperature.

$\frac{8 liters}{303 K}=\frac{16 liters}{T_{2}}$

$T_{2} = 606 K$

Convert this value back to Celsius.

C = K - 273 = 606 - 273 = 333^oC

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Example Question #2 : Gas Laws

Five moles of nitrogen gas are present in a 10L container at 30^oC. What is the pressure of the container?

Possible Answers:

17.2atm

5atm

9atm

12.4atm

10.5atm

Correct answer:

12.4atm

Explanation:

Using the ideal gas law equation we can find that P= nRT/V. We then plug in the given values.

$P=\frac{5mol*0.0821\frac{L*atm}{mol*K}*303K}{10L}$

Solving for P gives us 12.4atm.

Note: 30^oC must be converted into Kelvin by adding 273K

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Example Question #31 : Fluids And Gases

If one mole of oxygen gas occupies a 5L container at 300K, what is the pressure in the container?

$R=0.0821\frac{atm*L}{mol*K}$

Possible Answers:

4.9atm

15.0atm

10.0atm

1.0atm

7.0atm

Correct answer:

4.9atm

Explanation:

Using the ideal gas law, PV=nRT , we can solve for pressure.

$P(5L)= (1mol)(0.0821 \frac{atm*L}{mol *K})(300K)$

$P= \frac{(1mol)(0.0821\frac{atm*L}{mol*K})(300K)}{5L}=\frac{24.63atm*L}{5L}=4.9atm$

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Example Question #3 : Gases

Recall that the ideal gas law states that PV=nRT , where $\frac{L \cdot atm}{mol \cdot K}$ .

If there are 5g of hydrogen gas in a 10L chamber at 32°C, what would be the pressure?

Possible Answers:

12.5atm

3.15atm

6.25atm

1.3atm

Correct answer:

6.25atm

Explanation:

Using the equation PV=nRT and solving for P you get, $P = \frac{nRT}{V}$ .

Recall that hydrogen forms a diatomic molecule when in gas form. This should always be an assumption when working with hydrogen gas on the MCAT. When we convert 5g to moles, we must use a conversion factor of 2g/mol.

$5g\ H_2*\frac{1mol\ H_2}{2g\ H_2}=2.5mol\ H_2$

Temperature must be converted to Kelvin. You must have this conversion memorized for the MCAT.

T=32^oC+273K=305K

Now we can solve for P.

$P=\frac{nRT}{V}=\frac{(2.5mol)(0.0821\frac{L*atm}{mol*K})(305K)}{10L}=6.26atm$

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Example Question #1 : Gases

A certain gas is initially at a pressure of 2atm in a volume of 5L. It then experiences a decrease in volume to 2L, and is held at a constant temperature throughout the process. What is the new pressure?

Possible Answers:

5atm

3atm

4atm

2atm

Correct answer:

5atm

Explanation:

Since this is an isothermal change (constant temperature), this falls under Boyle's law.

P_1V_1 = P_2V_2

Taking this equation and solving for the new pressure (P₂) we come up with 5atm.

$P_2=\frac{P_1V_1}{V_2}=\frac{(2atm)(5L)}{(2L)}=5atm$

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Example Question #1 : Gases

A balloon filled with one mole of an ideal gas is leaking molecules at a constant rate of $\small0.01\frac{mol}{hr}$ .

After 75 hours, the pressure is half of the initial pressure. What is the new volume in terms of the initial volume, $\small V_i$ ?

Assume temperature remains constant.

Possible Answers:

2V_i

$\frac{1}{3}V_i$

$\frac{1}{5}V_i$

$\frac{1}{4}V_i$

$\frac{1}{2}V_i$

Correct answer:

$\frac{1}{2}V_i$

Explanation:

We can use the given values to determine how many moles of the gas have leaked out after 75 hours.

$0.01\frac{mol}{hr}*75hr=0.75mol$

We now know that the balloon started with one mole, and that 0.75 moles have leaked. This means that 0.25 moles remain in the balloon. We now have our initial and final mole values and pressure values. We can rearrange the ideal gas law to isolate the variables we need, assuming that the temperature is constant.

PV = nRT

$\frac{PV}{n}=RT=constant$

$\frac{P_1V_1}{n_1}=\frac{P_2V_2}{n_2}$

Using our proportions, we can try to solve for the final volume. For simplicity, assume the initial pressure is 1 and the final pressure is 0.5.

$\frac{(1)V_1}{1}=\frac{(0.5)V_2}{0.25}$

$V_1=2V_2\rightarrow V_2=\frac{1}{2}V_1$

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Example Question #7 : Gas Laws

How many moles of oxygen gas are in a $\small 560mL$ sample that is at a temperature of $\small 75^oC$ and a pressure of $\small 2.1atm$ ?

$R = 0.0821 \frac{L\cdot atm}{mol \cdot K}$

Possible Answers:

0.22mol

0.022mol

0.041mol

41mol

Correct answer:

0.041mol

Explanation:

First, each value must be converted to the correct units given for the gas constant.

$560 mL *\frac{1 L}{1000 mL} = 0.56 L$

75 ^oC +273 = 348 K

Next, use the ideal gas law to solve for moles.

PV = nRT

$n = \frac{PV}{RT}$

$n = \frac{(2.1 atm)(0.56 L)}{(0.0821)(348 K)} = 0.041 mol$

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Example Question #1 : Gas Laws

Regarding the following sets of conditions, which answer option gives the correct listing of systemic pressures from greatest to least?

$\text{I.}\ 2mol N_2,\ 3mol O_2;\ T=273K;\ V=10L$

$\text{II.}\ 5mol N_2;\ T=546K;\ V=5L$

$\text{III.}\ 2.5mol N_2;\ T=273K;\ V=10L$

$\text{IV.}\ 5mol N_2,\ 5mol O_2;\ T=546K;\ V=5L$

Possible Answers:

$P_{III}<P_{II}<P_I<P_{IV}$

$P_I=P_{III}<P_{IV}=P_{II}$

$P_{II}<P_{III}<P_{IV}<P_I$

$P_{III}<P_I<P_{II}<P_{IV}$

$P_{III}<P_I=P_{II}<P_{IV}$

Correct answer:

$P_{III}<P_I<P_{II}<P_{IV}$

Explanation:

This question asks for you to look at a set of conditions for gases, and determine relative pressures. The best equation to use for quick calculation and relation is the ideal gas law, given by:

PV=nRT

Rearranging this, and removing the constant (since it will not affect relative pressure), we can generate a proportionality of pressure to the other variables.

$P=\frac{nT}{V}$

We can use this proportionality with each option to determine their rankings by pressure.

$P_I=\frac{(5mol)(273K)}{10L}$

$P_{II}=\frac{(5mol)(546K)}{(5L)}=\frac{2(5mol)(273K)}{\frac{1}{2}(10L)}=4P_I$

$P_{III}=\frac{(2.5mol)(273K)}{10L}=\frac{\frac{1}{2}(5mol)(273K)}{10L}=\frac{1}2{}P_I$

$P_{IV}=\frac{(10mol)(546K)}{5L}=\frac{2(5mol)2(273K)}{\frac{1}{2}(10L)}=8P_1$

$\frac{1}{2}P_I<P_I<4P_I<8P_I$

$P_{III}<P_I<P_{II}<P_{IV}$

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Example Question #1 : Gas Laws

How many grams of ammonia gas are in a $\small 8.0L$ sample at $\small 30.0^oC$ and $\small 3.0atm$ ?

$R = 0.0821 \frac{L\cdot atm}{mol \cdot K}$

Possible Answers:

13g

32g

0.96g

16g

Correct answer:

16g

Explanation:

First, convert temperature to Kelvin.

30^oC+273=303K

Next, use the ideal gas law to solve for moles.

PV = nRT

$n = \frac{PV}{RT}$

$n = \frac{(3 atm)(8.0 L)}{(0.0821)(303K)} = 0.96 mol$

Finally, convert moles of ammonia to grams using molar mass.

$0.96mol\ NH_3*\frac{17.0g\ NH_3}{1mol\ NH_3}=16g\ NH_3$

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MCAT Physical : Gas Laws

Study concepts, example questions & explanations for MCAT Physical

All MCAT Physical Resources

Example Questions

Example Question #1 : Gas Laws

Example Question #1 : Gas Laws

Example Question #2 : Gas Laws

Example Question #31 : Fluids And Gases

Example Question #3 : Gases

Example Question #1 : Gases

Example Question #1 : Gases

Example Question #7 : Gas Laws

Example Question #1 : Gas Laws

Example Question #1 : Gas Laws

All MCAT Physical Resources

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